Question

find a and b if
(a+ib) (1+i)=2+i​

Answer 1

$(a + ib)(1 + i) = (2 + i) \\ = > a + ib = \frac{2 + i}{1 + i} \times \frac{1 - i}{1 - i} \\ = > a + ib = \frac{3 - i}{1 + 1} \\ = > a + ib = (\frac{3}{2} ) + ( \frac{ - 1}{2} )i$

Therefore, a = 3/2 and b = -1/2

Hope, it'll help you.....

Answer 2

Answer:

The values are a=\frac{3}{2}a=

2

3

and b=-\frac{1}{2}b=−

2

1

Step-by-step explanation:

Given : Expression (a+ib)(1+i)=2+i(a+ib)(1+i)=2+i

To find : The value of a and b ?

Solution :

Solve the expression LHS,

a(1+i)+ib(1+i)=2+ia(1+i)+ib(1+i)=2+i

a+ai+ib-b=2+ia+ai+ib−b=2+i

(a-b)+i(a+b)=2+i(a−b)+i(a+b)=2+i

Comparing LHS and RHS,

a-b=2a−b=2 ....(1) and a+b=1a+b=1 .....(2)

Solving these two equation by adding them,

a-b+a+b=2+1a−b+a+b=2+1

2a=32a=3

a=\frac{3}{2}a=

2

3

Substitute in equation (1),

\frac{3}{2}-b=2

2

3

−b=2

b=\frac{3}{2}-2b=

2

3

−2

b=-\frac{1}{2}b=−

2

1

Therefore, The values are a=\frac{3}{2}a=

2

3

and b=-\frac{1}{2}b=−

2

1